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Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

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GENERAL ORGANIC CHEMISTRY

C1 IUPAC System : Rules for a acyclic compounds :

(i) Saturated aliphatic hydrocarbons have been given the generic name alkanes.

Chain length Root Word Chain Length Root Word

C1

Meth– C8

Oct(a)

C2

Eth– C9

Non(a)

C3

Prop– C10

Dec(a)

C4

But(a)– C11

Undec(a)

C5

Pent(a)– C12

Dodec(a)

C6

Hex(a)– C13

Tridec(a)

C7

Hept(a)– C14

Tetradec(a)

(ii) A primary suffix : is added to the root word to designate the saturation or unsaturation in a carbon chainas shown below :

Hydrocarbon chain type Suffix Example

– Saturated –ane Methane

– Unsaturated with one – C = C – bond ene Ethene

– Unsaturated with one – C C bond yne Ethyne

(iii) A secondary suffix : Secondary suffix is added after dropping terminal ‘e’ of the primary suffix to indicatethe functional group present in a compound. The common secondary suffixes used to represent functionalgroups are :

Functional group Secondary suffix Examples

Alcohol (–OH) –ol Methanol

Aldehyde (– CHO) – al Methanal

Ketone ( = O) – one Propanone

Carboxylic acid (– COOH) – oic acid Propanoic acid

Carboxylic acid ester (– COOR) – oate Methyl ethanoate

Carboxylic acid chloride (– COCl) – oyl chloride Ethanoyl chloride

C2 Substituted Alkanes : The following rules are given by IUPAC to name substituted alkanes :

(1) Select longest chain of carbon atom

Consider the compound as derivative of the alkane containing that chain.

(2) Selecting and naming the parent chains : If there are two or more possibilities of longest carbon chain,the one with more substituent is preferred to the other

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(3) Numbering of carbon atoms in the parent chain : After selecting the longest chain preffered, numberthe parent carbon chain from the end which has a substituent nearest to it, so as to give minimum number tothe carbon atom bearing subtituent.

(4) Numbering the substituent present at equivalent positions in the parent chain : When two or moresubstituent are present at equivalent positions in the parent carbon chain, the lowest number is given to onethat comes first in the alphabetical order.

for e.g.,

‘E’ of Ethyl comes before ‘M’ of methyl substituent in alphabetical order.

NOTE : Group like –Cl, –Br, –I, –NO2, –N = O and –F, C

6H

5– are also considered as substituents and not

as functional groups and are also numbered in alphabetical order. e.g.

– If alphabetical order is to be followed for groups like isopropyl (i) neopenthyl(n), first letter (i) and (n) isconsidered but for sec-butyl and tert-butyl only B is considered. e.g.

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(5) Lowest sum rule : If two substituents are at equivalent positions from both ends and a third substituent isalso present, the numbering is given from the end nearest to both the substituents so as to give the lowestsum number possible to side chain bearing carbon. e.g.

sum of position of substituent

C3 Procedure to write the names of alkanes : Arrangement of prefixes, root word and suffixes. These arearranged as follows while writing the name

(substituent name) : Prefixes acc. to alphabetical order + root word + primary suffix + secondary suffix.

e.g. (i)

(Substituent with carbon no.) Prefix 3-methyl

Root word pent

Primary suffix ane

(ii)

Prefix 3-methyl

Root word hex

Primary suffix ane

(iii) (iv)

(v)

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(vi)

(vii)

(viii)

(ix) (x)

(xi)

(xii)

[Branched side chain is also numbered from the point of attachement to parent chain as 3and2,1 etc. to

distinguish it from the numbering of parent carbon chain]

II. Alkenes, Alkynes and Substituents :

For selecting and numbering such a continuous carbon chain, a double or triple bond ia always given apriority over the side chain/substituent and is included in the chain even if it voilates the longest chain rule.

It double bond and triple bond is present at equivalent position then double bond is given preference

e.g. 2

1234

HCHCCHC . But if triple bond is present on the one end of carbon chain and double bond is

present on the second carbon on the other end of carbon chain then triple bond is preferred. e.g.

3

654

2

321

HCHCHCHCCCH

General order of preference is : Double bond > Triple bond > Substituents.

When more than one double or triple bond are present then di, tri or tetra words are used with the numericalbefore root word followed by proper suffix to indicate two, three and four double or triple bondsrespectively. e.g.,

Butene12

12

2

3

3

4

HCHCHCHC

Butene2

3

123

3

4

HCHCHCHC

enechloroprop3

2

32

2

1

lCHCHCHC

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[Ethyl is given preference over methyl substituent]

[both the double bonds are included in parent chain although it is not

the longest chain]

– When alkane, alkene and alkyne are substituents then they are named as alkyl, alkenyl and alkynylrespectively.

e.g. : allyl

22propyl

223ethynylethenyl

2ethyl

23 CHCHCH,CHCHCH,CHC,CHCH,CHCH

– When these groups are substituents that is in side chain, the carbon linked to parent carbon chain is given

number 1 . If side chain contain double or triple bond then they are not numbered according to priority e.g.

(III) Compounds having functional groups : The following rules are observed while naming suchcompounds :

(i) Maximum number of functional groups are included in the selected carbon chain even if it violates thelongest chain rule. The parent chain is so selected so as to give the lowest number to the carbon atombearing functional groups.

(ii) If functional group(s) bearing carbon atom in itself (e.g., – COOH, –CHO, –CN etc.) is (are) linked in thecontinuous carbon chain, such carbon should also be numbered.

(iii) Halo, nitro (–NO2), nitroso (–NO) are considered as substituents not as functional group.

(iv) Substituents are prefixed alphabetically before the parent hydrocarbon without considering the presence offunctional groups.

(v) The suffix representing the functional group is added at the end by dropping ‘e’ of the correspondinghydrocarbon. For functional groups like nitrile ‘(e)’ should not be dropped e.g. CH

3CN is named as

Ethanenitrile.

(vi) Functional groups like (–NH2), alkylamine (RNH–), dialkylamino (R

2N–), carbylamino )CN(

and alkoxy

(RO–) are used as prefixes.

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(vii) The carbon number indicating the position of the functional group is inserted either before the name ofhydrocarbon or by dropping the ‘e’ of the hydrocarbon followed by the functional group. e.g.

(viii) In compounds containing more than one identical functional groups, the word di-, tri- terta- are used withtheir numerals for two, three and four identical groups respectively. In such cases numerals are writteneither before or after the name of hydrocarbon without dropping ‘e’ followed by suffix of the functionalgroup. e.g.,

(ix) If double bond, triple bond and substituents are also present then preference order is :

Functional group > Double bond > Triple bond > Substituent

Following examples illustrate the above rules :

(i) (ii)

(iii) (iv)

(v) (vi)

(vii) (viii)

(ix) (x)

(xi) (xii)

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(xiii) (xiv)

(xv) (xvi)

(xvii) (xviii)

(xix) (xx)

(xxi) (xxii)

(xxiii)

– When groups like –COOH, CHO are present at terminal ends then there is no need to number them as 1.

– If two such functional groups are present at terminals and also substituents in remaining chain, select thatend which can give minimum number to alphabetically preferred substituents also. Similarly double ortriple bond is preferred to substituents.

(xxiv) (xxv)

Polyfunctional Compounds :

When different functional groups are present then the group which comes first in the following priorityorder is called as principal functional group and is given priority in numbering the carbon chain :

Group Prefix name Suffix name

– COOH – carboxy – oic acid

– SO3H – sulpho – sulphonic acid

– COOR – carbalkoxy – oate

– COX – haloformyl – oylhalide

– CONH2

– carboxamide – amide

– CN – cyano – nitrile

– CHO – formyl – al

C = O – oxo orketo – one

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– OH – hydroxy – ol

– SH –Mercapto – thiol

– NH2

– amino – amine

– OR – alkoxy –

– O – – epoxy –

– N = N – – azo –

– C = C – – enyl – ene

– C C – – ynyl yne

– While selecting the principal chain it should include maximum number of functional groups and rest offunctional group will act as substituent.

Principal functional group > Double bond > Triple bond > substituents.

The following examples illustrate the above guidelines

(i)

(ii) (iii)

(iv) (v)

(vi)

IUPAC Nomenclature of cycloalkenes :

(1) Cycloalkanes : are alkanes in which carbon atoms are arranged in a ring. These are named by adding theprefix cyclo to the name of the alkane having the same number of carbon atoms as in the ring.

– These are represented by simple geometric figures as :

– Substituted cycloalkanes are named as alkyl cycloalkanes.

– The numbering of carbon atoms in the ring is done in such a way that the substituent which comes first inthe alphabetical order is given the lowest possible number provided it does not violate the lowest sum oflocants rule. e.g.

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– When the ring has more carbon atoms than the alkyl group attached to it or equal no. in both the cases alkylgroup is treated as substituent

– In case alkane chain contain greater number of carbon atoms than ring, then ring is treated as substituent.

– Cycloalkanes and Cycloalkynes :

In case of substituted cycloalkenes and cycloalkynes :

– Alicyclic compounds containing functional group :

Alicyclic alcohols, aldehydes, ketones, acids etc. are named in the same fashion as corresponding aliphaticcompounds by prefixing the word cyclo before the name.

– If side chain contains the multiple bond or a functional group, the alicylic ring is treated as substituentirrespective of the size of the ring.

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Naming of Aromatic Compounds

Some derivatives of benzene have universally accepted, common names (written in brackets below) whichhave no resemblance to the names of the attached substituents.

Disubstituted of trisubstituted benzenes are named by numbering the positions of the substituents.

Dimethylbenzenes are commonly called xylenes.

If different substituents are attached to the benzene ring then the following rules are observed.

(a) The principal group of substituent of the base compound is assigned the number 1.

(b) The numbering is so done that the substitution are located at the lowest numbers possible.

(c) The substitutions are written in alphabetical order.

(d) In some cases the common name of the benzene derivative is considered as the name of the parentcompound.

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Let us now look at a few categories of benzene derivatives.

The trivial names phenyl, benzyl, benzal and benzo have been accepted by IUPAC.

Halogen derivatives

Hydroxy derivatives

Amines

Ketones

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Aromatic ethers

Aldehydes

Carboxylic acids

Organic compounds having same molecular formulae but differing from each other at least in somephysical or chemical properties or both are known as isomers.

C4 ISOMERISM : There are two mains types of isomerism :

(i) Structural isomerism

(ii) Space or stereoisomerism

Chain Isomerism : This type of isomerism is due to difference in the arrangement of carbon atom consti-tuting the chain i.e., straight or branched chain of carbon atoms.

For e.g., n-Butane and Iso-butane are chain isomer.

i.e., CH3CH

2CH

2CH

3 and

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Position Isomerism : This arises due to the difference in the positions occupied by particular atoms orgroups (substituents) in the same carbon chain or due to different position of double or triple bonds in thealkenes and alkynes.

For e.g., C3H

8O

Functional Isomerism : Compounds having the same molecular formulae but different functional groupsshow functional isomerism.

As the functional group largely determines the properties of compound, such isomers differ in theirphysical and chemical properties.

For e.g.,

(1) Alcohols and Ethers (CnH

2n+2O)

(2) Aldehyde, Ketones & Unsaturated Alcohols (CnH

2nO)

(3) Carboxylic Acids and Esters (CnH

2nO

2)

(4) Oximes and Amides (CnH

2n+1NO)

(5) 10, 20 and 30 amines (CnH

2n+3N)

(6) Alkynes and Alkadiene (CnH

2n–2)

(7) Cyanides and Isocyanides

(8) Nitroalkanes and Alkylnitrites

Ring and Chain Isomer : This type of isomerism is due to different mode of linking of carbon atoms i.e.,the isomers possess either open or closed chain isomers.

For e.g.,

(1) Alkenes and Cyclo Alkanes (CnH

2n)

(2) Alkynes and Cyclo Alkenes (CnH

2n–2)

Tautomerism : This is a special type of functional isomerism where the functional isomers exist inequilibrium with each other. The two forms which are in equilibrium in each other are called tautomers.

For e.g.,

(1) Keto & Enol are tautomeric forms.

(2) Nitro and Aci

(3) Oximes and Nitroso

Enolisation is in order as follows :

CH3CHO < CH

3COCH

3 < CH

3COCH

2CHO < CH

3COCH

2COCH

3

Metamerism : It is the isomerism in the same homologous series. It is due to presence of different allkylgroups attahced to the same polyvalent functional group or atom.

For e.g.,

(1) C4H

10O C

2H

5OC

2H

5C

3H

7OCH

3

Diethylether Methylpropyl ether

(2) C5H

10O C

2H

5COC

2H

5C

3H

7COCH

3

Diethyl ketone Methylpropyl ketone

(3) C4H

11N C

2H

5NHC

2H

5C

3H

7NHCH

3

Diethylamine Methylpropyl amine

Practice Problems :

1. The number of isomers of C6H

14 is

(a) 4 (b) 5 (c) 6 (d) 7

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2. An isomer of ethanol is

(a) methanol (b) acetone

(c) diethyl ether (d) dimethyl ether

3. The compound C4H

10O can show

(a) metamerism (b) position isomerism

(c) functional isomerism (d) all the three

4. C7H

8O shows how many isomers (having onebenzene ring).

(a) 2 (b) 3 (c) 4 (d) 5

5. Keto-enol tautomerism is observed in :

(a) (b)

(c) (d) (CH3)

3CCHO

6. Maximum enolisation takes place of

(a) CH3COCH

3(b) CH

3COCH

2CHO

(c) CH3COCH

2COCH

3(d) = O

[Answers : (1) b (2) d (3) d (4) d (5) b (6) d]

C5 Stereoisomerism :

They are of two types :

(1) Geometrical isomerism : Such isomers, which possess the same molecular and structuralformula but differ int he arrangement of atoms or groups in space due to hindered rotation

around the double bond i.e. ( )

The isomers which has similar group on same side are called -cis isomer and similar groups onopposite sides are called as -trans isomer.

However in case of all the four groups are different, for e.g., ( ), it is not possible

to decide -cis and -trans isomers. In such cases E and Z system of nomenclature is used.

(2) Optical isomerism : The compounds having similar physical and chemical properties butdiffering only in the behaviour towards the plane polarised light are called as optical isomers.

On the basis of study of optical activity, the various organic compounds are divided into threetypes :

(a) The optical isomer with rotate the plane of polarised light to the right are called asdextrorotatory or d-form or indicated by + sign.

(b) The optical isomer which rotate the plane or polarised right to the left are called asleavoratotory or l-form on indicated by –ve sign.

(c) The optical powers of the above two isomers are same i.e. of same magnitude but ofpositive sign.

(d) The equimolar mixture of two forms, therefore will be optically inactive. This mixtureis termed is Racemic Mixture.

Cause of Optical Activity or Optical Isomerism :

A plane which divides an object into two symmetrical halves is said to have a plane of symmetry. An objectlacking a plane of symmetry is called dissymmetric or chiral.

A symmetric object is called as achiral.

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A dissymmetrical or chiral object can be defined as the one which is not super imposable on its mirrorimage. In order to exhibit optical activity an object must be chiral.

The compounds which are mirror images of each other but not super imposable are called as enantiomers.

Generally compunds having carbon atom to which four different groups are when attached showoptical isomerism. Molecule having a symmetric carbon atom lacks in plane of symmetry. Thus compoundwhich consist of at least one asymmetric carbon atom is capable of showing the phenomenon of opticalisomerism.

Exception : Following compound with a cumulative system of double bond and having no chiral carbonatom shows optical isomerims.

but not the following compound

i.e., .

Number of possible stereoisomers in compounds containing different numbers of asymmetriccarbons :

(i) When the molecule has no symmetry : n ® number of asymmetric carbon atoms.

The number of d- and l-forms a = 2n

The number of meso forms m = 0

The total number of optical isomer = a + m = 2n

(ii) When the molecule has symmetry :

n number of asymmetric carbon atoms which is even.

The number of d- and l-form, a = 2n – 1

The number of meso forms, m = 1

2

n

2

The total number of optical isomer = a + m = 1

2

n1n 22

.

(iii) When the molecule has symmetry and n no. of asymmetric carbon atom which is odd :

The number of d and l-forms, a =

2

1n

1n 22

The number of meso forms 2

1n

2m

.

Total optical isomer = a + m = 2n – 1.

Diastereomers : Stereoisomers which are not mirror images of each other are called diastereomers.

Practice Problems :

1. The molecular formula of a saturated compound is C2H

4Cl

2. This formula permits the existance of

two

(a) functional isomers (b) position isomers

(c) optical isomers (d) cis-trans isomers

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2. Number of optically active isomers of tartaric acid

CH(OH)COOH |CH (OH)COOH is :

(a) 2 (b) 3 (c) 4 (d) 5

[Answers : (1) b (2) a]

GENERAL ORGANIC CHEMISTRY

C6A Reagents :

Following types of reagents are used

(i) Electrophiles : Electrophilic reagents (electron loving) are those reagents that are positivelycharged or contain an electron deficient atom. i.e., whose octet is not complete. They are alsoknown as Lewis Acid (they are electron acceptor). For e.g., NO

2+, Cl+, H+, SO

3, BF

3 etc.

(ii) Nucleophiles : Nucleophilic reagents (nucleus loving) or electron repelling reagents arenegatively charged or having neutral atom with lone pair. They are also called as Lewis base(electron pair donor). For e.g., OH–, CN–, NH

3, H

2O etc.

C6B Types of Intermediates Formed in the Organic Reaction :

(i) Carbocation or Carbonium Ion : The carbon with a positive charge and six electron pair

in valence shell is known as carbocation. This carbon is sp2 hybridised .

The carbocation can undergo rearrangemets reaction.

Order of Stability of Carbocation : Benzyl, Allyl > 30 > 20 > 10 > CH3+.

(ii) Carbanion : The carbon with a negative charge having three bond pair and one lone pair iscalled as carbanion. This carbon is sp3 hybridised.

Order of stability of carbanion is : Benzyl, Allyl > CH3– > 10 > 20 > 30.

(iii) Free Radical : The carbon containing single unpaired electron is called as carbon free radical. It

is sp2 hybridised

Order of Stability of Free Radical : Benzyl, Allyl > 30 > 20 > 10.

(iv) Carbene : There is another class of intermediate called carbene, (methylene). It exists in

two forms :

(a) Singlet Carbene : In which the unshared electrons are paired. i.e., . It is sp2 hybridised

(b) Triplet Carbene : In which unshared electrons are not paired. i.e., . It is sp hybridised

(v) Nitrene

(vi) Benzyne

Practice Problems :

1. Most-stable carbonium ion is

(a)2462 HCHCNOp

(b)256 HCHC

(c)246 HCHCClp

(d)2463 HCHCOCHp

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2. Arrange the following in the decreasing order of the free radical

(a) II > I > III (b) I > II > III (c) III > II > I (d) I > III > II

[Answers : (1) b (2) a]

C7 Different Effects for the Displacement of Shared Pair of Electron are as follows :

(i) Inductive Effect (ii) Resonance Effect

(iii) Electromeric Effect (iv) Hyperconjugation Effect

(i) Inductiv Effect : Electrone withdrawing and electron releasing group is attached to the carbonchain permanent polarity is induced on the carbon atom. This displacement of electrons towardsone of the atom because of difference in electronegativity is known as inductive effect.

This effect is of two types :

(a) Positive Inductive Effect (+I) : This is due to electron releasing group. It develops anegative charge on the carbon chain.

+I effect decreases as we move away from the electron releasing group.

Greater the size of alkyl group greater the electron releasing effect. Due to electronreleasing group electron density increases hence basic nature increased, thus acidicdecreases

(b) Negative Inductive Effect (–I) : This is due to electron withdrawing groups. It developespositive charge on the chain.

–I effect also decreases as we move away from the electron withdrawing group. Afterthird carbon, this effect is negligible. –I effect is in the order as follows :

NO2 > F > COOH >Cl > Br > I > OH > C

6H

5

Due to –I effect electron density decreases hence basic nature is decreased and acidicnature is increased.

(ii) Resonance Effect : The phenomena by which two or more structures involving identicalpositions of atoms can be written for a particular compound is called Resonance. The variousstructures are called as resonating structures.

Resonating structure can be written for the following :

(a) Presence of conjugated system (Alternate double-single bond or alternatesingle-multiple bonds) for e.g.,

(b) Conjugated system attached to electron deficient atom with vacant p-orbital, for e.g.,

(c) Conjugated system between electron rich atom i.e., containing lone pair and alternate

double bond for e.g.,

(In this case Cl is having –I as well as +R effect)

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Resonance effect is of two types :

(a) +R Effect : Groups which can released or donate the electron pairs through resonanceare called as +R effect. for e.g., OH–, OCH

3–, NH

2– etc.

(b) –R Effect : Groups which attrach or withdraw the electron through resonance are of

–R effect for e.g., NO2–, CN–, , –COOH etc.

(iii) Electromeric Effect : It is a temperory effect, it comes into play instantaneously at the demandof attacking reagent and as soon as the attaching reagent is removed the original conditionsrestored.

The effect involving the complete transfer of shared pair of -electrons to one of the atomsjoined by the multiple bond (double or triple) at the requirement of attacking reagent is known aselectromeric effect. It is of two types :

(a) +E Effect (b) –E Effect

(iv) Hyperconjugation Effect : It is the conjugation between and bond, for e.g.,

This type of electron released by alkyl group attached to the unsaturated group is called hyperconjugation. There should be atleast one H atom at carbon atom w.r.t. (C = C) bond.

Practice Problems :

1. Delocalised molecular orbitals are found in

(a) H2

(b) HS– (c) CH4

(d) CO3

2–

2. The correct order of the increasing C – O bond length of CO, CO32–, CO

2 is

(a) CO32– < CO

2 < CO (b) CO

2 < CO

32– < CO

(c) CO < CO32– < CO

2(d) CO < CO

2 < CO

32–

3. Among the following which one is most basic

(a) NH3

(b) CH3NH

2

(c) CH3CH

2NH

2(d) C

6H

5NH

2

4. Most acidic compound is

(a) CH3COOH (b) C

6H

5COOH (c) O

2NC

6H

4CO

2H (d) C

6H

5OH

5. In the following compounds phenol (I), p-cresol (II), m-nitrophenol (III) and p-nitrophenol (IV), theorder of acidity is

(a) III > IV > I > II (b) I > IV > III > II

(c) II > I > III > IV (d) IV > III > I > II

6. The most unlikely representation of resonance structures of p-nitrophenoxide ion is

(a) (b) (c) (d)

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7. Increasing values of dissociation constant Ka of

CH3COOH (I), HCOOH (II), – COOH (III) :

(a) I < II < III (b) III < II < I (c) I < III < II (d) II < III < I

8. Give the order of the stabilities of the followings :

(a) b > a > c > d > e (b) a > b > c > d > e

(c) e > b > a > c > d (d) d > a > b > c > e

9. The acylium ion, the structure R — C

O : is more stable than :OCR..

(a) As in R — C

O : octet of every atom is complete.

(b) In R — C

O : structure there are moree covalent bonds than :OCR..

(c) Both are correct

(d) Given statement is wrong as :OCR..

is more stable than R — C

O :

[Answers : (1) d (2) d (3) c (4) c (5) d (6) c (7) c (8) a (9) c]

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1. The IUPAC name of picric acid is

(a) 2, 4, 6-trinitrophenol

(b) 2, 4, 6-trinitro-1-hydroxy hexane

(c) 2, 4, 6-trinitro-1-hydroxy benzene

(d) 1, 3, 5-trinitro-6-hydroxy benzene

2. Vinyl alcohol and acetaldehyde are

(a) geometrical isomers

(b) keto-enol tautomers

(c) chain isomers

(d) position isomers

3. Number of isomers of molecular formula C2H

2Br

2

is

(a) 1 (b) 2

(c) 3 (d) 4

4. Which of the following compounds exhibits stere-oisomerism ?

(a) 2-methylbutene-1

(b) 3-methylbutyne-1

(c) 3-methylbutanoic acid

(d) 2-methylbutanoic acid

5. Which of the following hydrocarbons has thelowest dipole moment ?

(a)

(b) CH3C CCH

3

(c) CH3CH

2C CH

(d) CH2 = CH – C CH

6. Polarisation of electrons in acrolein may be writ-ten as

(a)

OCHCHHC 2

(b) OHCCHHC 2

(c) OHCHCHC 2

(d)

OHCHCHC 2

7. Which of the following is not an electrophile

(a) AlBr3

(b) BF3

(c) SnCl4

(d) NH3

SINGLE CORRECT CHOICE TYPE

8. Markownikoff’s rule is applicable to which of thefollowing reactions

(a) C2H

4 + HBr (b) C

3H

6 + Cl

2

(c) C3H

6 + HBr (d) C

3H

6 + Br

2

9. Which of the following statements is correct abouta carbonium ion ?

(a) it reacts with nucleophile

(b) it can undergo rearrangement

(c) it can eliminate an H+ to form an olefin

(d) all are correct

10. What is the decreasing order of strength of the bases

OH—(I), NH2

— (II), H — C C— (III),CH

3CH

2— (IV) ?

(a) IV > II > III > I (b) III > IV > II > I

(c) I > II > III > IV (d) II > III > I > IV

11. The intermediate during the addition of HCl topropene in presence of peroxide is

(a) CH3 — HC

— CH2Cl

(b) CH3 — HC

— CH3

(c) CH3 — CH

2 — HC

2

(d) CH3 — CH

2 — HC

2

12. In the following compounds, anisole (I), benzene(II) and nitro benzene (III), the case of reaction withelectrophiles is

(a) II > III > I (b) III > II > I

(c) II > I > III (d) I > II > III

13. Among the given compounds, the most susceptibleto nucleophilic attack at the carbonyl group is

(a) MeCOCl (b) MeCHO

(c) MeCOOMe (d) MeCOOCOMe

14. The IUPAC name of the compound having formula

(a) 3-amino-hydroxy propionic acid

(b) 2-amino-propan-3-oic acid

(c) Amino hydroxy propanoic acid

(d) 2-amino-3-hydroxy propanoic acid

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New Delhi – 110 018, Ph. : 9312629035, 8527112111

15. The IUPAC name of the compound,

is

(a) 2-Bromo-3-chloro-4-oxopentanoic acid

(b) 3-Chloro-2-bromo-4-oxopentanoic acid

(c) 4-Carboxybromo-3-chloro-2-butanone

(d) None

16. The IUPAC name of the compound,

(a) 1, 3-Dibromo-3-methylbutane

(b) 3-Methyl-1, 2-bromobutane

(c) 3-Methyl-1, 3-bromopropane

(d) None

17. 6 g of the organic compound on heating with NaOHgave NH

3 which is neutralised by 200 mL of 1 N

HCl. Percentage of nitrogen is

(a) 12% (b) 60%

(c) 46.67 % (d) 26.67%

18.

F,OH,HN,HC 23 in increasing pKb values

are :

(a)

FOHHNHC 23

(b) 32 CHNHOHF

(c) FCHNHOH 32

(d) none is correct

19. B.E. of C — H bonds designated by I, II, III, IV,

V in increasing order is :

(a) V < IV < III < II < I

(b) III < II < IV < V < I

(c) III < II < V < IV < I

(d) all equal

20. Stability order of ...... in order :

– 2HC

, CH2 = CH – 2HC

I II

C)CH( 33 HCCH2

III IV

(a) IV < III < II < I (b) IV < II < I < III

(c) I < II < III < IV (d) IV < I < III < II

21. Which shows aromatic character ?

(a) (b)

(c) (d)

22. Degree of unsaturation in

is

(a) 2 (b) 3

(c) 4 (d) 5

23. The electronegativity of cesium is 0.7 and that offluorine is 4.0. The bond formed between the two is

(a) covalent (b) electrovalent

(c) coordinate (d) metallic

24. In which of the following molecules the bond angleis maximum ?

(a) CH4

(b) H2O

(c) NH3

(d) CO2

25. Total number of line pair of electrons in XeF4 is

(a) 0 (b) 1

(c) 2 (d) 3

26. CO2 is isostructural with

(a) SO2

(b) HgCl2

(c) C2H

2(d) SnCl

2

27. Which of the following is the most polar bond ?

(a) Cl – Cl (b) N – F

(c) C – F (d) O – F

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28. (A) Tetracyanomethane

(B) Carbon dioxide

(C) Benzene

(D) 1, 3 butadiene

Ratio of and bonds is in order

(a) A = B < C < D (b) A = B < D < C

(c) A = B = C = D (d) C < D < A < B

29. In sp3d hybridisation, the d orbital that participatesin hyubridisation is

(a) dx2 – y2 (b) dz2

(c) dxy (d) dxz

30. C2 – C

3 sigma single bond in vinyl acetylene is due

to the overlapping of

(a) sp – sp (b) sp2 – sp2

(c) sp – sp2 (d) sp2 – sp

31. In the compound of the type ECl3, where E = B, P,

As or Bi the angles Cl – E – Cl for different E are inthe order

(a) B > P = As = Bi (b) B > P > As > Bi

(c) B < P = As = Bi (d) B < P < As < Bi

32. A -bonds is formed by the overlap of

(a) s-s orbital

(b) s-p orbital

(c) p-p orbital in end to end fashion

(d) p-p orbital in sidewise manner

33. Mg2 C

3 reacts with water and forms propyne C

34–

which has

(a) two sigma and two pi - bond

(b) three sigma and one pi - bond

(c) two sigma and one pi - bond

(d) two sigma and three pi - bonds

34. Element X is strongly electropositive and Y isstrongly electronegative. Both are univalent. Thecompound formed would be

(a) X+ Y– (b) X – Y

(c) X– Y+ (d) X Y

35. The correct order of the O – O bond length inO

2, H

2O

2 and O

3 is

(a) O3 > H

2O

2 > O

2(b) O

2 > H

2O

2 > O

3

(c) O2 > O

3 > H

2O

2(d) H

2O

2 > O

3 > O

2

36. The molecule which has highest dipole momentamongts the following is

(a) CH3Cl (b) CH

2Cl

2

(c) CHCl3

(d) CCl4

37. Pairs of related chemical species are given below.Which pair is not related through resonance ?

(a) CH3 – CO – CH

2 – COOC

2H

5, CH

3 –

C(OH) = CH – COOC2H

5

(b)

(c)

(d)

38. Which is maximum basic in nature ?

(a)

(b)

(c)

(d)

39. Give the decreasing order of acidic characters ofthe following :

(i) (ii)

(iii) (iv)

(a) I > II > III > IV (b) III > IV > I > II

(c) II > III > IV > I (d) I > III > IV > II

40. Arrange the following according to their stability

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New Delhi – 110 018, Ph. : 9312629035, 8527112111

(a) II > IV > I > III > V > VI

(b) II > IV > I > III > V > VI

(c) I > II > III > IV > V > VI

(d) VI > V > IV >III > II > I

41. The 1, 2 addition product of free radical additionof CBrCl

3 to 1, 3-butadiene

(a)

(b)

(c)

(d) None

42. Arrange in decreasing order of basicity.

m - NO2 - C

6H

4 - NH

2C

6H

5NH

2

I II

p - NO2 - C

6H

4NH

2o - NO

2 - C

6H

4NH

2

III IV

(a) II > I > III > IV (b) I > II > III > IV

(c) IV > III > II > I (d) IV > II > III > I

43. Among 3-Butenoic acid and 3-Butynoic acid thegreater acidic character is

(a) 3-Butenoic acid

(b) 3-Butynoic acid

(c) both

(d) none

44. Basic character of following in the decreasing or-der is :

(i) (CH3)

3N (ii)

(iii) CH3 – C N

(a) iii > ii > i (b) i > ii > iii

(c) ii > i > iii (d) i > iii > ii

45. Acidic character of the following in the increasingorder is :

(i) (ii)

(iii)

(a) iii < ii < i (b) ii < i < iii

(c) i < ii < iii (d) ii < iii < i

46. Arrange the following compounds in order of in-creasing b.pts.

(i) CH3COCl (ii) (CH

3CO)

2O

(iii) CH3CONH

2(iv) CH

3COOH

(a) CH3COCl < CH

3COOH < (CH

3CO)

2O <

CH3CONH

2

(b) CH3COCl > CH

3COOH = (CH

3CO)

2O =

CH3CONH

2

(c) CH3COCl = CH

3COOH > (CH

3CO)

2O >

CH3CONH

2

(d) CH3COCl < CH

3COOH = (CH

3CO)

2O >

CH3CONH

2

ANSWERS

(SINGLE CORRECT CHOICE TYPE)

1. a

2. b

3. c

4. d

5. b

6. a

7. d

8. c

9. d

10. a

11. b

12. d

13. a

14. d

15. a

16. a

17. c

18. a

19. c

20. a

21. b

22. c

23. b

24. d

25. c

26. c

27. c

28. a

29. b

30. d

31. b

32. d

33. a

34. a

35. d

36. b

37. a

38. b

39. c

40. a

41. a

42. a

43. b

44. b

45. c

46. a

CGOC– 24


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E X E R C I S E(SUBJECTIVE)

1. Write the tautomeric structures of :

(i) Phenol (ii) AAE or EAA

(iii) 1,3,5-cyclohexatrione

(iv) AA

(Lactam-Lactim tautomerism)

(v) B (Ketimide-

enamine tautomerism)

(vi) CH3 – N = O C (Nitroso-

isonitroso tautomerism)

(vii) D

(Nitro-isonitroso tautomerism)

(viii) H – C N E (Cyanuric-

isocyanuric acid tautomerism)

2. Arrange the following in the order of decreasingbasic character

(a) (1) CH3NH

2

(2) (CH3)

2NH

(3) C6H

5NH

2

(4) NH3

(b) (1) C6H

5NH

2

(2) p-NO2.C

6H

4.NH

2

(3) m-NO2.C

6H

4NH

2

(4) p-CH3O.C

6H

4.NH

2

(c) (1) p-toluidine

(2) N, N-dimethyl-p-toluidine

(3) p-nitroaniline

(4) aniline

(d) (1) Methylamine

(2) Dimethylamine

(3) Aniline

(4) N-Methylaniline

3. Arrange in decreasing order of basicity.

(a) Acidic character of :

(i) 3-Butenoic acid

(ii) 3-Butynoic acid

(b) Basic character of :

(i) (CH3)

3N

(ii)

(iii) CH3 – C N

(c) Acidic character of :

(i) (ii)

(iii)

4. Arrange the following compounds in order ofincreasing b.pts.

(i) CH3COCl (ii) (CH

3CO)

2O

(iii) CH3CONH

2(iv) CH

3COOH

5. (a) K1 for Maleic acid is greater than

Fumaric acid OR pK1 for Maleic acid is

less than Fumaric acid. Explain it ?

(b) K2 for Fumaric acid is greater than

Maleic acid OR pK2 for Fumaric acid is

less than Maleic acid. Explain it ?

6. Give the decreasing order of acidic or basiccharacters of the following :

(a) Acidic Character

(i) (ii)

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(iii) (iv)

(b) Acidic character :

(i) (ii)

(iii) (iv)

(c) Basic character :

(i) (ii)

(iii)

7. Why cyanoanilium ion is

stronger acids than anilinium ion.

8 Give the order of acidic character, o, p- &m-cyanoanilinium ions.

9. The pka of p – F – C6H

4COOH is 4.14 whereas that

of p – Cl – C6H

4COOH is 3.99.

10. Give the order of acidic character of the following :

(i) (a) p-Nitrobenzoic acid

(b) 2,4-Dinitrobenzoic acid

(c) 2, 4, 6-Trimethyl benzoic acid

(d) 2, 3, 4-Trinitrobenzoic acid

(ii) 2, 6-Di-tert-butyl phenol is much weakeracid than phenol.

(iii) (a) phenol (b) Cyclohexanol(c) p-Bromophenol (d) p-Methoxyphenol

(iv) (a) Benzoic acid (b) p-tert-butylbenzoicacid (c) p-Nitrobenzoic acid

11. Give the order of the stabilities of the followings :

(i)

(ii)

(iii)

12. Why pyridine is much weaker base than aliphaticamines.

13. Arrange the following in decreasing order of dipolemoment.

(i)

(ii) (iii)

14. Which has greater Ka value ?

(i)

(ii)

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15. Give the order of the stabilities of the following freeradicals.

(a)

(b)

(c)

(d)

(e)

(d)

16. Why are mercaptans (Ka 10–111) more acidic than

alcohols (Ka 10–17) ?

17. Acidic character

(a) (i) Benzene

(ii) CH3OH

(iii) H2O

(iv) CH3SH

(b) (i) CH3CH

2COOH

(ii) HOCH2CH

2COOH

(iii) ClCH2CH

2COOH

(iv) O2NCH

2CH

2COOH

(c) (i)

(ii)

(iii)

(iv)

(d) (i) HCHO

(ii) CH3CHO

(iii) PhCHO

(iv)

(v)

(vi) (CH3CO)

2O

(vii) RCOOR

(viii) RCONH2

18. Name each of the following compounds :

(a)

(b)

19. (a) Arrange the following anions indecreasing order of basicity :

HO—, NH2—, —:C N:, NO

3—

(b) Arrange the following group in order ofincreasing basicity :

3, 4-Dichloroaniline, 4-Chloro-2-nitroaniline, 4-Chloro-3-nitroaniline

(c) For nitromethane molecule, writestructure(s)

(i) showing significant resonancestablisation

(ii) indicating tautomerism

20. Assign the A.I.U.P.A.C. name for the followingcompounds :

(i)

(ii)

21. Arrange the following in order of increasing basicnature.

(a) diphenyl amine, aniline, cyclohexyl amine

(b) pyridine, pyrolle,aniline

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New Delhi – 110 018, Ph. : 9312629035, 8527112111

ANSWERS (SUBJECTIVE)

1. (i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

2. (a) 2 > 1 > 4 > 3 (b) 4 > 1 > 3 > 2

(c) 2 > 1 > 4 > 3 (d) 2 > 1 > 4 > 3

3. (a) ii > i (b) i > ii > iii

(c) i > ii > iii

4. CH3COCl < CH

3COOH < (CH

3CO)

2O < CH

3CONH

2

5. (a) Both F.A. and M.A. have +ve ionisable hydrogen. Maleate ion is stabilised byintramolecular H-bonding, so M.A. is stronger than F.A. or K

1 for M.A. is greater

than K1 of F.A. or pK

1 for M.A.is less than pK

1 of F.A., as shown below :

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6. (a) i > ii > iii > iv

(b) iii > i > ii > iv

(c) i > ii > iii (steric inhibition of resonance)

7. The reactions are :

The –CN gp is withdrawing because of resonance and inductive effect. This explains the stron-ger acidity of cyanoanilinium ion.

8. o > p > m, The ortho cyano derivative is the stronger acid because of the resonance (–R) & strongestinductive effect (–I). In para-cyano compounds only resonance effect equal to that of ortho isomer)is operative. At para inductive effect (–I) is very very weak.

9. In +R & –I effect nearly cancel each other, whereas in +R effect is veryy

small, due to the size of 2p & 3p orbitals of F & Cl, respectively, which they offer for +R effect.

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10. (i) b > d > a > c

(ii) Due to steric inhibition of resonance of bulky o-tert-butyl groups and this also preventsolvation of the corresponding phenoxide ion which makes it less stable than phenoxideion. Hence the observed effect.

(iii) c > a > d > b (iv) c > a > b

11. (i)

(ii)

(iii) a > b > d > e > c > f

12. Pyridine is weaker base than R NH2, due to +I effect in aliphatic amines.

13. i > ii > iii. In case of (i), intramolecular H-bonding decreases bond angle (< 600) hence dipole momentis increased. In case of (iii), dipole-dipole repulsion of two Cl atoms in ortho-position increases bondangle (> 600) hence dipole moment is decreased.

14. i > ii

15. a > b > d > f > c > e

16. 1. The S atom is larger and is more polarizable than O atom. S compounds are morepowerful nucleophile and compounds containing SH gps are stronger acid than theiroxygen analogs C

2H

5S— ion stronger nucleophile than C

2H

5O— ion.

2. Bond dissociation energy of S – H bonds of thiols (~ 330 kJ) is much less than that of(O – H) bond of alcohols (~ 420 kJ). Thiols undergoes oxidative coupling reaction withMild O.A. Alcohols do not undergoes such reaction.

2RS – H + H2O

2 RS – SR (Disulfide) + 2H

2O

In alcohols the oxidation takes place a the weaker C – H (~ 360 kJ) bond rather than at the strongerO — H bond.

17. (a) iii > iv > ii > i (b) iv > iii > ii > i

(c) iv > i > ii > iii

(d) i > ii >iii > iv > v > vi > vii > viii

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